Alice is exploring the wonderland, suddenly she fell into a hole, when she woke up, she found there are b - a + 1 treasures labled a fromb in front of her.
Alice was very excited but unfortunately not all of the treasures are real, some are fake.
Now we know a treasure labled n is real if and only if [n/1] + [n/2] + ... + [n/k] + ... is even.
Now given 2 integers a and b, your job is to calculate how many real treasures are there.
Input
The input contains multiple cases, each case contains two integers a and b (0 <= a <= b <= 263-1) seperated by a single space. Proceed to the end of file.
Output
Output the total number of real treasure.
Sample Input
0 2 0 10
Sample Output
1 6
一开始还考虑用欧拉定理算因子和什么的。。。
然后看到数据范围。。。噗。。显然不能那样做。。。
所以一开始看清楚数据范围是很重要的。
这个数据范围应该就是找规律了。。。。
结果试着找了下。。。果然有规律hhhhh
需要注意的是要分奇偶。。。
如果是偶数可能会多算。。。因为这个wa了一发。。。
1 /************************************************************************* 2 > File Name: code/zoj/3629.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年10月24日 星期六 20时07分02秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<iomanip> 10 #include<cstdio> 11 #include<algorithm> 12 #include<cmath> 13 #include<queue> 14 #include<vector> 15 #include<stack> 16 #include<cctype> 17 18 #define yn hez111qqz 19 #define j1 cute111qqz 20 #define ms(a,x) memset(a,x,sizeof(a)) 21 using namespace std; 22 const int dx4[4]={1,0,0,-1}; 23 const int dy4[4]={0,-1,1,0}; 24 typedef long long LL; 25 typedef double DB; 26 const int inf = 0x3f3f3f3f; 27 LL a,b; 28 void pre() 29 { 30 int sum = 1; 31 for ( int i =1; i <= 300 ; i++) 32 { 33 int res = 0 ; 34 for ( int j = 1 ; j <= i ; j++) 35 res = res + i/j; 36 if (res%2==0) 37 { 38 sum++; 39 cout<<"i:"<<i<<endl; 40 } 41 42 } 43 44 printf(" sum: %d ",sum); 45 46 } 47 48 LL cal( LL n) //计算0到n有多少个数满足 49 { 50 if(n==0) return 1; 51 if (n<0) return 0; 52 53 LL res = 0 ; 54 LL sn = sqrt(n); 55 LL k = sn/2 + 1; //k为等差数列的项数，第k项为4k-3 56 57 res = (2*k-1)*k; 58 if (sn%2==0) 59 { 60 res = res -(4*k-3); 61 res = res+n-(2*k-2)*(2*k-2)+1; 62 } 63 64 // cout<<"k:"<<k<<" n:"<<n<<" res:"<<res<<endl; 65 return res; 66 67 } 68 int main() 69 { 70 #ifndef ONLINE_JUDGE 71 freopen("in.txt","r",stdin); 72 #endif 73 74 //pre(); 75 while (scanf("%lld %lld",&a,&b)!=EOF) 76 { 77 printf("%lld ",cal(b)-cal(a-1)); 78 } 79 80 81 #ifndef ONLINE_JUDGE 82 fclose(stdin); 83 #endif 84 return 0; 85 }